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            <h1 style="display: none">report-牛客CSP-S提高组赛前集训营4</h1>
            
              <p class="note note-info">
                
                  本文最后更新于：2 年前
                
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            <div class="markdown-body" id="post-body">
              <h3 id="0-x-前言"><a href="#0-x-前言" class="headerlink" title="[0.x]前言"></a>[0.x]前言</h3><p><strong><a target="_blank" rel="noopener" href="https://ac.nowcoder.com/acm/contest/1103">牛客CSP-S提高组赛前集训营4</a></strong></p>
<p>语言：$Cpp$</p>
<a id="more"></a>
<h3 id="目录"><a href="#目录" class="headerlink" title="目录"></a>目录</h3><ul>
<li>[0.x]前言</li>
<li>[1.x]A.复读数组</li>
<li>[2.x]B.路径计数机[消化ing]</li>
<li>[3.x]C.排列计数机[消化ing]</li>
</ul>
<hr>
<h3 id="1-x-A-复读数组"><a href="#1-x-A-复读数组" class="headerlink" title="[1.x]A.复读数组"></a>[1.x]A.复读数组</h3><h4 id="题目描述："><a href="#题目描述：" class="headerlink" title="题目描述："></a><strong>题目描述</strong>：</h4><p>有一个长为$n<em>k$的数组，它是由长为$n$的数组$A_1,A_2,…,A_n$<em>*重复k次</em></em>得到。</p>
<p>定义这个数组的一个区间的权值为它里面<strong>不同的数的个数</strong>，现在，你需要求出对于这个数组的每一个非空区间的权值之和。</p>
<p>答案<strong>对$10^9+7$取模</strong>。</p>
<h4 id="被教育历史："><a href="#被教育历史：" class="headerlink" title="被教育历史："></a><strong>被教育历史</strong>：</h4><p><del>没有思路被GOD胡教育</del></p>
<p><del>推不出算式被GOD胡教育</del></p>
<p><del>把<strong>iterator</strong>写成指针被GOD胡教育</del></p>
<p><del><strong>没有取模</strong>没过样例被GOD胡教育</del></p>
<p><del><strong>没开longlong</strong>见祖宗被GOD胡教育</del></p>
<h4 id="题目分析："><a href="#题目分析：" class="headerlink" title="题目分析："></a><strong>题目分析</strong>：</h4><p>此题如果从扫描的角度去做肯定做不出了，要从<strong>贡献</strong>考虑。由于蒟蒻对于这一类题十分不熟悉所以完全没有思路。<del>(就是菜)</del></p>
<p>由于数列重复得到，所以我们肯定期望能在一个循环节里就得出答案，所以我们考虑一个循环节中每一个数对答案的<strong>贡献</strong>。</p>
<p><img src="https://s2.ax1x.com/2019/11/06/MPtaOs.png" srcset="/blog/img/loading.gif" alt=""></p>
<p>对于第一个循环中(图中深灰色)的数字$X$，我们发现，包含$X$且满足$X$是区间内第一个$X$(维护它是“不同的数”的性质)的左边界和右边界分别可以在下面的区域中取：</p>
<p><img src="https://s2.ax1x.com/2019/11/06/MPNV7q.png" srcset="/blog/img/loading.gif" alt=""></p>
<p>因此我们可以发现，符合条件并且可以维护性质的区间有$len<em>橙*len</em>蓝$个</p>
<p>当我们处理第二个$X$的时候我们会发现，这次蓝色区间依然是到末尾，但是橙色区间不可以，因为如果橙色区间包括了之前的那个$X$那当前的$X$就不是第一个$X$了，也就是说此时它<strong>不再贡献</strong>，所以这次左右边界分别在下面区间取得。</p>
<p><img src="https://s2.ax1x.com/2019/11/06/MPUxW8.png" srcset="/blog/img/loading.gif" alt=""></p>
<p>同样，符合条件的有$len<em>橙*len</em>蓝$个</p>
<p>这是第一个循环节的情况，仅仅对于<strong>非首</strong>的数，每一个循环节内的<strong>贡献计算</strong>都是类似的。但是首位的X，至少第一个循环和之后的循环不类似，同时也与非首位的计算不类似。</p>
<p>但是怎么算呢，显然我们需要有每一个循环节内同一个数字的位置，但我们又需要该循环节内所有数字的信息，第一反应是桶，但显然需要离散化，然后还得在桶里塞一个支持随机访问的数据类型。虽然蒟蒻对于STL很生疏，但是还是硬着头皮写上了<code>map&lt; int , vector&lt;int&gt; &gt;</code></p>
<p>现在我们用$now[i]$来表示某个数字的第$i+1$个的位置(从$now[0]$开始存)，那么可以得到，当$i\geq 1$时，对于<strong>第一个循环节</strong>，当前<strong>贡献</strong>为：</p>
<script type="math/tex; mode=display">
(now[i]-now[i-1])*(k*n-now[i]+1)</script><p>写出之后的几个循环节以后进行简单的等差数列求和和化简得到，对于<strong>所有的循环节</strong>，当前<strong>贡献</strong>为：</p>
<script type="math/tex; mode=display">
(now[i]-now[i-1]) * (\frac{(k+1)*k}{2}*n-now[i]*k+k)</script><p>同理对刚刚说明的需要特判的首数进行讨论，可以得到余下公式。</p>
<h4 id="AC代码"><a href="#AC代码" class="headerlink" title="AC代码"></a>AC代码</h4><div class="hljs"><pre><code class="hljs cpp"><span class="hljs-meta">#<span class="hljs-meta-keyword">include</span> <span class="hljs-meta-string">&lt;bits/stdc++.h&gt;</span></span>
<span class="hljs-keyword">using</span> <span class="hljs-built_in">std</span>::<span class="hljs-built_in">cin</span>;<span class="hljs-keyword">using</span> <span class="hljs-built_in">std</span>::<span class="hljs-built_in">cout</span>;<span class="hljs-keyword">using</span> <span class="hljs-built_in">std</span>::<span class="hljs-built_in">endl</span>;<span class="hljs-keyword">using</span> <span class="hljs-built_in">std</span>::<span class="hljs-built_in">memset</span>;<span class="hljs-keyword">using</span> <span class="hljs-built_in">std</span>::<span class="hljs-built_in">map</span>;
<span class="hljs-keyword">typedef</span> <span class="hljs-keyword">long</span> <span class="hljs-keyword">long</span> ll;
<span class="hljs-keyword">typedef</span> <span class="hljs-keyword">unsigned</span> <span class="hljs-keyword">long</span> <span class="hljs-keyword">long</span> ull;
<span class="hljs-keyword">long</span> <span class="hljs-keyword">long</span> n,k,x,ans;
<span class="hljs-keyword">const</span> ll p = <span class="hljs-number">1e9</span> + <span class="hljs-number">7</span>;
<span class="hljs-built_in">map</span>&lt;<span class="hljs-keyword">int</span> , <span class="hljs-built_in">std</span>::<span class="hljs-built_in">vector</span>&lt;<span class="hljs-keyword">int</span>&gt; &gt; mp;
<span class="hljs-function"><span class="hljs-keyword">int</span> <span class="hljs-title">main</span><span class="hljs-params">()</span></span>&#123;
    <span class="hljs-built_in">cin</span> &gt;&gt; n &gt;&gt; k;
    <span class="hljs-keyword">for</span>(<span class="hljs-keyword">int</span> i = <span class="hljs-number">1</span>;i &lt;= n;++i)&#123;
        <span class="hljs-built_in">cin</span> &gt;&gt; x;
        mp[x].push_back(i);
    &#125;
    <span class="hljs-keyword">for</span>(<span class="hljs-built_in">map</span>&lt;<span class="hljs-keyword">int</span> , <span class="hljs-built_in">std</span>::<span class="hljs-built_in">vector</span>&lt;<span class="hljs-keyword">int</span>&gt; &gt;::iterator super = mp.begin();super != mp.end();++super)&#123;
        <span class="hljs-built_in">std</span>::<span class="hljs-built_in">vector</span>&lt;<span class="hljs-keyword">int</span>&gt; &amp;now = super-&gt;second;
        <span class="hljs-keyword">for</span>(<span class="hljs-keyword">int</span> i = <span class="hljs-number">1</span>;i &lt; now.size();++i)&#123;
            ans+=(now[i]-now[i<span class="hljs-number">-1</span>]) * ((((k+<span class="hljs-number">1</span>)*k/<span class="hljs-number">2</span>)%p)*n-now[i]*k%p+k)%p;
            ans%=p;
        &#125;
        ans%=p;
        ans += (now[<span class="hljs-number">0</span>]) * (n*k%p - now[<span class="hljs-number">0</span>] + <span class="hljs-number">1</span>)%p;<span class="hljs-comment">//特判 </span>
        ans%=p;
        ans += (now[<span class="hljs-number">0</span>]+n-now.back()) * ((((k<span class="hljs-number">-1</span>)*k/<span class="hljs-number">2</span>)%p)*n%p-now[<span class="hljs-number">0</span>]*(k<span class="hljs-number">-1</span>)%p+(k<span class="hljs-number">-1</span>))%p;
        ans%=p;
    &#125;
    <span class="hljs-built_in">cout</span> &lt;&lt; (ans%p+p)%p;
    <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;
&#125;</code></pre></div>
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